[Www.vietmaths.com]Ung Dung Bat Dang Thuc Tim Gtln Va Gtnn
Transcript of [Www.vietmaths.com]Ung Dung Bat Dang Thuc Tim Gtln Va Gtnn
Vn dng bt ng thc tm GTLN - GTNN v gii phng trnh
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PHN M U
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Vn dng bt ng thc tm GTLN - GTNN v gii phng trnh
I. L DO CHN TI C th ni trong chng trnh ton bc trung hc ph thng th phn kin thc v bt ng thc l kh kh. Ni v bt ng thc th c rt nhiu bt ng thc c cc nh Ton hc ni ting tm ra v chng minh. i vi phn kin thc ny th c hai dng bi tp l chng minh bt ng thc v vn dng bt ng thc gii cc bi ton c lin quan. L mt sinh vin ngnh ton ti khng ph nhn ci kh ca bt ng thc v mun tm hiu thm v cc ng dng ca bt ng thc phc v cho vic ging dy ton sau ny. Do ti chn ti Vn dng bt ng thc t m gi tr ln nht, gi tr nh nht v gii phng trnh tm hiu thm. Khi vn dng bt ng thc gii cc bi ton dng ny th c rt nhiu bt ng thc chng ta vn dng. y ti ch gii hn trong ba bt ng thc l bt ng thc Csi, Bunhiacopski v bt ng thc vect. Trong ti ny ti trnh by cch v n dng ba bt ng thc trn tm gi tr ln nht, gi tr nh nht v gii phng trnh rn luyn kh nng vn dng bt ng thc gii ton v qua c th tch ly c kinh nghim trong gii ton ging dy sau n y.
III. I TNG NGHIN CU
i tng ca ti l ba bt ng thc Csi, Bunhiacopski v bt ng thc vect cng vi cc bi ton tm gi tr ln nht, nh nht v cc phng trnh. ti ny ch yu xoay quanh ba i t ng trn bn cnh ti cng gii thiu v chng minh mt s bt ng thc thng d ng khc.
Phm vi ca ti ny ch xoay ch yu vo ba bt ng thc nu trn gii cc bi ton tm gi tr ln nht, nh nht v gii phng trnh. VI. PHNG PHP NGHIN C U
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Tm v tham kho ti liu, su tm phn tch v bi tp gii minh ha, tham kho kin ca cn b hng dn
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IV. PHM VI NGHIN CU
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Mc tiu chnh ca ti ny l tng hp cc bi ton tm gi tr ln nht, nh nht v gii phng trnh bng bt ng thc ch yu vn dng ba bt ng thc ni trn. Qua y ti hi vng s a ra y cc dng vn ca cc bt ng thc ni trn.
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II. MC CH NGHIN CU
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Vn dng bt ng thc tm GTLN - GTNN v gii phng trnh
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PHN NI DUNG
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Vn dng bt ng thc tm GTLN - GTNN v gii phng trnh
Phn 1: S LC V BT NG THC
1.2. Tnh cht c bn ca bt ng thc
1.3.1. Bt ng thc cha tr tuyt i a b a b du = xy ra
ie t
1.3. Mt s bt ng thc c bn
m
a b a c b c a c b c a b a b c a c b a b c d a c e b e f a b v m 0 ma a b v m 0 ma a b 0 c d 0 ac bd an bn a b 0 a b
dmb mb
f
n
a
tab 0
a
b
a b
a1 a2 ... an n n a1.a2 ...an n Du = xy ra a1 a2 ... an M rng: Cho n s dng a1 , a2 ,..., an n c: 1 ... n 1 . Th: 2
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a1 a2 ... an a1 a2 ... an 1.3.2. Bt ng thc Csi Cho hai s dng a, b ta c: a b 2 ab Du = xy ra a b Tng qut: cho n s khng m a1 , a2 ,..., an n
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h2 , ta lun c: 2 v n s
s
.c o1
Cho hai s thc a, b bt k, ta nh ngha: a b a b 0
m,2
1.1. nh ngha bt ng thc
,....,
n
dng
a1 1 .a2 2 ...an n 1a1 Du = xy ra a1 a2
a ... ... an2 2
n n
a
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Vn dng bt ng thc tm GTLN - GTNN v gii phng trnh
1.3.3. Bt ng thc Bunhiacopski Bt ng thc Bunhiacopski: Cho hai b s a, b v c, d ta c: 2 ac bd a2 b2 c2 d 2 a b Du = xy ra c d Tng qut: Cho n s a1 , a2 ,..., an v b1, b2 ,.., bn ty ta c:
a1b1 a2b2 ... anbnDu = xy ra
2
a12a2 b2 ...
2 2 2 2 a2 ... an b12 b2 ... bn
a1 b1
an bn
a1a2 ...am a1m
b1b2 ...bm
c1c 2 ...cm
m
m
Cho a 1 v r N : n Nu n 1 th 1 a
1 na du = xy ran
a
Du = xy ra a1 : b1 : ... : c1 a2 : b2 : ... : c2 1.3.4. Bt ng thc Bernuolli
t
m b1m ... c1m a2
m m m b2 ... c2 ... am
h
M rng: Cho m b s, mi b gm n s khng m: ai , b i ,...c i Khi ta c:
sw
.c oa
m m bm ... cm
... an : bn : ... : cn
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u vu v
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Nu a n 1 th 1 a 1.3.5. Bt ng thc vect
1 na
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mi 1, 2,..., m 0 hoc n 1
Vn dng bt ng thc tm GTLN - GTNN v gii phng trnh
Phn 2: TM GI TR LN NHT V GI TR NH NHT CA HM S HOC BIU THC
2.1 KIN THC CN NH
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i vi vic tm gi tr ln nht, gi tr nh nht ca biu thc (hm s) th c th k n cc phng php sau: phng php kh o st, phng php nh gi thng thng v phng php s dng bt ng thc. Trong cc ph ng php nu trn th phng php s dng cc bt ng thc c th coi l mt trong nhng phng php thng dng v hiu qu nht tm gi tr ln nht v nh nht ca biu thc v hm s. i vi phng php ny, ta s dng cc bt ng thc thng dng nh: bt ng thc Csi, Bunhiacopski, Schwartz, Bernouli, bt ng thc vect nh gi biu thc P (h oc hm s f ( x1, x2 ,..., x n ) ), t suy ra gi tr ln nht, gi tr nh nht cn t m. Phng php ny, nh tn g i ca n, da trc tip v o nh ngha ca gi tr ln nht v nh nht ca biu thc v hm s. Lc chung ca phng php ny c th miu t nh sau: - Trc ht chng minh mt bt ng thc c dng P ( x1 , x2 ,..., xn ) D vi bi ton tm gi tr nh nht (hoc P ( x1 , x2 ,..., xn ) D i vi bi ton tm gi tr ln nht), y P l biu thc hoc hm s xc nh trn D. - Sau ch ra mt phn t ( x01 , x02 ,..., x0 n ) D sao cho P( x01 , x02 ,..., x0 n ) . Ty theo dng ca bi ton c th m ta chn mt bt ng thc thch hp p dng vo vic tm gi tr nh nht v ln nht. Do phm vi ca ti, y ch gii thiu phng php s dng ba bt ng thc l: Csi, Bunhiacopski v phng php b t ng thc vect.
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2.1.2. Tm gi tr nh nht, gi tr ln nht ca biu thc (h m s) bng phng php vn dng bt ng thc
t1 x3
Cho biu thc P( x1, x2 ,..., xn ) ( hm s f ( x1, x2 ,..., x n ) ), xc nh trn D ( x1, x2 ,..., xn ) D v - Nu P( x1, x2 ,..., x n ) M (hoc f ( x1, x2 ,..., xn ) M ) ( x1, x2 ,..., x n ) D sao cho: P( x1, x2 ,..., xn ) M th M gi l gi tr ln nht ca P( x1, x2 ,..., xn ) (hoc f ( x1, x2 ,..., x n ) ). K hiu l maxP hoc Pmax ( max f ( x1, x2 ,..., x n) hoc f ( x1, x2 ,..., x n )max ). - Nu P( x1, x2 ,..., x n ) m ( hoc f ( x1, x2 ,..., x n ) m ) th m gi l gi tr nh nht ca P( x1 , x2 ,..., xn ) ( hm s f ( x1, x2 ,..., x n ) ). K hiu l minP hoc Pmin (min f ( x1, x2 ,..., x n ) hoc f ( x1 , x2 ,..., x n )min ).
h55
s1 2 x 33
V d : Tm gi tr nh nht ca hm s f ( x) x 2 Gii: Ta c: f ( x)1 2 x 3 1 2 x 3 1 2 x 3 1 x3
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2.1.1. nh ngha
2 ( x 0) x35 ( BT Csi) 27
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Vn dng bt ng thc tm GTLN - GTNN v gii phng trnh
Du = xy ra Vy Min f x =5
1 2 x 35
1 x3
x5
3
x
5
3
5 ti x 27
3
2.2. BI TP 2.2.1. S dng bt ng thc Csi
p dng bt ng thc Csi ta c:
1
a b
2
a b1
1
b c
m
a
Gii:
2
b c8 abc abc 8
t
Bi 1: Cho ba s thc dng a, b, c . Tm gi tr nh nht ca biu thc sau: a b c P 1 1 1 b c a
h12
s1111 1 a
Lu : bit c bi ton no s dng bt ng Csi ta cn ch n cc thnh phn ca hm s hoc biu thc. Nu n c dng tch hoc l tng ca hai phn khng m v c bit sau khi vn dng bt ng thc Csi th xut hin biu thc ca gi thit ban u v a c v hng s th ta c th s dng bt ng thc Csi nh gi tm gi tr ln nht, nh nht.
.c oc a 22.1b
Suy ra Hay
a b c 1 1 b c a
P 8
Du = xy ra khi v ch khi a b c 1 Vy Pmin
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Bi 2: Cho ba s thc a, b, c 0 tha
1
1 a 1 b 1 cabc
Tm gi tr ln nht ca biu thc M
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Gii:
Ta c:
11 1 a
1
11 1 b
1 a 1 b 1 c1
21
1 1 a1 1 c
1 b 1 cc 1 b 1 c
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Vn dng bt ng thc tm GTLN - GTNN v gii phng trnh
p dng bt ng thc Csi ta c:
b
c
1 b 1 c
2
bc 1 b 1 c
1 1 a
2
bc 1 b 1 c
(1)
1 1 b 1 1 c
2
2
ab 1 a 1 b
.c o s8 a 2b 2 c 2 2 2 1 a 1 b 1 c2
ac 1 a 1 c
m1 1 c
Tng t, ta c:
(2)
(3)
T (1) , (2) v (3) nhn v vi v ta c:
1 1 a
1 1 b
1 a 1 b 1 cSuy ra:
M
abc
1 8
Du = xy ra khi v ch khi
11 8
1
m1
aa b c 1 2(tha iu kin ban u)
1 a 1 b 1 cVy M max Cch khc: ti a b c
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1 2
T gi thit ta c:
1 b 1 c
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1 a 1 c
t8
1
1 a 1 b
h
abc 1 a 1 b 1 c
21 a 1 b 1 c
2 a b c
3 ab bc ac 2 1 a 1 b 1 c(1)
p dng bt ng thc Csi ta c:
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2abc ab bc ac
4 4 2a 3b 3c 3
(2)
T (1) v (2) ta c: 1 4 4 2a3b3c3
1 8abc hay M
abc
1 8
Du = xy ra khi v ch khi 2abc ab bc ac
a b c
1 2
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Vn dng bt ng thc tm GTLN - GTNN v gii phng trnh
Vy M max =
1 8
ti a b c
1 2
Bi ton tng qut:i
Tm gi tr ln nht ca biu thc M Lp lun nh trn ta c M max
.c oa2
a1.a2 ....an
m... an 1 n 1(1) (2) (3) (4) (5) (6)
Cho a1 , a2 ,..., an
0 tha mn :
n
1 1 1 ai
n 1
2 n ti a1
Gii: p dng bt thc Csi ta c:4
a1 x 2
1 x2
4
1 x .4 1 x
4
1 x 1 x
4
1 x .1 1 x .1
1 x 1 2 1 x 1 2
4
4
T (1), (2), (3) cng v theo v ta c:
Nhn thy (4) xy ra khi v ch khi (1), (2) v (3) ng thi xy ra khi v ch khi x 0 .
Li p dng bt ng thc Csi, ta c:
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f ( x) 1
1 x
ie t1 x
m1 x 2
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1 x
1 x .1
1
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1 x
1 x .1
1
1 x 2
T (5), (6) a n:
1 x
1 x 2
t1 xx D
h1
xc nh trn D
x
R : 1 x 1 . Tm gi tr ln nht ca f ( x) trn D.
s
Bi 3: Cho hm s f ( x)
4
1 x2
4
1 x
4
1 x
1 x
1 x 3
(7)
Du = (7) xy ra khi v ch khi (5) v (6) ng thi xy ra khi v ch khi x 0 . T (4) v (7) suy ra f ( x) 3
x D.
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Vn dng bt ng thc tm GTLN - GTNN v gii phng trnh
Ta li c f (0) 3, v 0 D . Do : max f ( x) = 3.
Bi 4: Tm gi tr nh nht ca hm s thc sau:
f ( x)
vi
0 x 1
Ta c: f ( x)
1 1 x 1 x
1 x x x 1 x
1
1 x 1 x
1 x x 2 x 1 xp dng bt ng thc Csi ta c:
f ( x)
Bi 5: Cho ba s thc dng a, b, c .
m
Vy min f ( x) 4 ti x
1 2
a
Du = xy ra khi v ch khi
1 x x
tx 1 x a
1 x x 1 x x 2 2 . x 1 x x 1 x
h
s2 4x 1 2 b c a c a b b c x y z 2
Tm gi tr nh nht ca biu thc P Gii: t:
x b c,
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y c a,
z
a b
a b cV
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1
x y z
2
a
y z x , 2
b
z x y , 2
.c ox
Gii:
m1 1 x x x
1 1 x 1 x
c
(*)
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T ta c:
P
y z x 2x
z x y 2y
x y z 2z
1 y z 2 x
z x y
x y z
3
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1 2
y x
x y
z x
x z
z y
y z
3
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Vn dng bt ng thc tm GTLN - GTNN v gii phng trnh
1 2 2 2 3 2
3 2
( Bt ng thc Csi)
Du = xy ra
T (*) ta c a b c Vy Pmin
Bi 6: Cho ba s thc dng a, b, c tha: a b c 1 . Tm gi tr ln nht ca biu thc S abc a b b c c a Gii:
p dng BT Csi cho ba s dng, ta c:
a b c 33 abcV
m
1 3 3 abc
a
t(1)
ie t
a b
b c
c a
33 a b b c c a(2)
2 33 a b b c c aT (1) v (2) nhn v vi v ta c:
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2 9 3 abc a b b c c a8 93 S
S
8 729 1 3
Du = xy ra khi v ch khi a b c
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Vy
Smax
8 729
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Bi 7: Tm gi tr ln nht ca hm s f ( x ) trn min D
x
R: 1 x
1 . 2
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s
3 vi mi s thc dng a, b, c tha a b c . 2
1 x 2 x2
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x y x z y z
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Vn dng bt ng thc tm GTLN - GTNN v gii phng trnh
Gii: Nhn thy D l min xc nh ca f ( x) . p dng bt ng thc Csi ta c:
1 x 2x2 x 1 2
1. 1 x 2 x 2 1 x 2x2 2
m
1
1 x 2 x2 2
x D
Do :
f ( x)
f ( x) 1 x 2T suy ra:
f ( x) 1
x D
Mt khc du = xy ra th
1 1 x 2x2 1 x2 1 xTa li c: f (0) 1 Vy max f ( x) 1x D
t1 1 22 1 . x0
hx 0 D1 1 x2
Bi 8: Cho hm s f ( x)
Tm gi tr nh nht ca f ( x) vi x Gii:
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Ta c: f ( x )
1 x
ie t
1 x
2
2
1 x2
m1 x2
2 1 x
p dng bt ng thc Csi ta c:2
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f ( x)
1 2 x .2 x
16
Du = xy ra
x 1 > 0.
Vy min f ( x ) 16 ti x 1x 0
Bi 9: Cho ba s thc dng a, b, c . Tm gi tr nh nht ca biu thc sau: Trang 12
a
1 x
s
.c o
Vn dng bt ng thc tm GTLN - GTNN v gii phng trnh
A
abc 1
1 aGii:
1 b
1 c
a b
b c
c a
a b c
A
ab
p dng bt ng thc Csi ta c:
ab
a b
2a ,
bc
b c
2b , 1 aa 1 c 1 a 6
ac 1 b 1 c
c a
T suy ra: A
2a 2b 2c1 a 2 b.a b
A
a b c 1 a
a
A 2 a.
1 bb
2 c.
Vy MinA = 6 ti
a
c 1
Bi ton tng qut:Cho P
ie t
a1.a2 ...an 1
1 a1
m
Du = xy ra
c 1
1 a2
...
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a1 a2 .a3 ...an
a2 a1.a3 ...an
...
t1 anc3 b3
1 b
1 c
hb
s1 b1
a b cc 1 c
(BT Csi)
an a1.a2 ...an
vi ai 0
i 1, n Th MinP = 2n ti a1
a2
... an 11 ab 2 1 bc 2 1 ca 2 c3 a3 b3 0, b 0, c 0 v abc 1
Tm gi tr nh nht ca P vi a
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Bi 10: Cho biu thc sau: P
a 3 b3 c 3
Gii:
Ta c: P
3
a3 b3
a3 c3
b3 c3
b3 a3
c3 a3
Trang 13
.c o2c
a b
bc
b c
ac
c a
1 a
1 b
1 c
a b c
a1 a2 ... an
m
Ta vit biu thc A li di dng sau:
Vn dng bt ng thc tm GTLN - GTNN v gii phng trnh
a 4b 2 c3
ab 5 c3
b 4c 2 a3
bc 5 a3
a 5c b3
a 2c 4 b3
ab 2 bc 2 ca 2
(1)
p dng bt ng thc Csi ta c:
m6
a3 b3
a3 c3
b3 c3
b3 a3
c3 a3
c3 b3
66
a3 a3 b3 b3 c3 c3 . . . . . b3 c3 c3 a 3 a 3 b3
(2)
66
a 4b 2 c3
ab5 c3
b 4c 2 a3
bc5 a3
a 5c b3
a 2c 4 b33abc 3
.c o6n
a 4b 2 c3
ab 5 c3
b 4c 2 a3
bc 5 a3
a 5c b3
a 2c 4 b3
a 4b 2 ab 5 b 4c 2 bc 5 a 5c a 2c 4 . . . . . c3 c3 a3 a3 b3 b3
6abc
(3) (4)
ab 2 bc 2 ca 2
3 3 ab 2 .bc 2 .ca 2
T (1), (2), (3) v (4) ta c:P 3 6 6 3 18
Du = xy ra
a
b
c 1
Vy Pmin = 18 ti a b c 1
Bi 11: Cho n s dng x1 , x2 , x3 ,..., xn
m
an 2 tha mn x1
tx2 ... xn
h
s
1
Gii: t a
ie t
a a Tm gi tr ln nht ca biu thc S x1a1 .x2 2 ...xn n , Trong : a1 , a2 , a3 ,..., an l n s dng cho trc.
w .v
a1 a2 ... an ,b1 b2
bibn
ai a
i 1, 2, .., n th bi
0
V b1 b2 ... bnx1 a1 x . 2 a2
1 . p dng bt ng thc Csi m rng ta c:x ... n an b1 x1 a11 aa a a1a1 .a2 2 ...an n
b b2 x2 ... n xn a2 anx1 x 2 .. x 1 a
wS
a a x1a1 .x2 2 ...xn n
Du = xy ra
1 aa x1 a1 1 a
w
xn x1 x2 ... xn x1 x2 x2 ... a2 an a1 a2 ... an a1 a2 xn ai x1 x2 ... xi i 1, 2,..., n a1 a2 an a
...
xn an
Vy Smax
1 a1 a2 an a1 .a2 ...an aa
Trang 14
Vn dng bt ng thc tm GTLN - GTNN v gii phng trnh
P
Gii: p dng bt ng thc Bunhiacopski ta c:2
3 4 a b c
h
1. 4a 3 1. 4b 3 1. 4c 3
1 1 1 4a 3 ab 3 ac 39
tP 4a 3 4b 3 4c 3Du = xy ra
3 4.3 93 7
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4a 3 4b 3 1 1 a b c 1
a
s4c 3 12
.c o63
Bi 1: Cho a, b, c
3 v a b c 3 . Tm gi tr ln nht ca biu thc sau: 4 4a 3 4b 3 4c 3 .
a
Vy MinP = 3 7 ti a
w
Bi 2: Cho cc hng s dng a, b, c v cc s dng x, y , z thay i sao cho a b c 1 . Tm gi tr nh nht ca biu thc A x y z . x y z Gii: a b c Ta c: a b c x y z x y z p dng bt ng thc Bunhiacopski ta c:
w .v
ie t
a , b, c
3 4
b
c 1.
2
w
a
b
c2
a x xx y z
b y y
c z z
a x
mb
2.2.2. S dng bt ng thc Bunhiacopski Lu : p dng c bt ng thc Bunhiacopski th hm s hoc biu thc hoc cc biu thc gi thit phi c dng tch ca hai biu thc hoc tng ca cc biu thc m chng l tch ca hai tha s. V sau khi p dng bt ng thc Bunhiacopski th phi c phn a v biu thc gi thit ban u v a c v hng s.
c 1
b y
c z
x
y z
a
b
c
Du = xy ra
a x x
b y y
c z z
a x
b y
c z
(1)
Trang 15
Vn dng bt ng thc tm GTLN - GTNN v gii phng trnh
Mt khc:
a x
b y
c z
1a b a a b b c
(2)
T (1) v (2) suy ra: x
z2
c
a
b
c
Vy maxA =
a
b
c
Bi 3: Tm gi tr nh nht ca hm s f ( x, y , z ) x 4 y 4 z 4 , trn min D x, y , z : x , y , z 0 v xy yz zx 1 Gii:
p dng bt ng thc Bunhiacopski ta d c:
t
1.x 2 1. y 2 1.z 2 xyV xy
2
3 x4 y2z2
y4 z22
z4 x2
h
s(1)
a
Li p dng bt ng thc Bunhiacopski ta c:
.c ox2 y2 z22
yz
zx
2
x2x2 y2
y2
z2
m
yz
zx 1 nn:
m(2)yz zx 1
y
c
1
ie ty z x
T (1) v (2) ta c: 3 x 4
y4
z4
1
w .vy z1 3
1 3 Du = xy ra khi v ch khi (1) v (2) ng thi xy ra x2 y 2 z 2 x y z kt hp vi iu kin xy f ( x, y , z )
Ta c: x
3 3
Vy Max f ( x, y , z )( x, y,z ) D
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Bi 4: Cho cc s dng a, b, c tha a 2 b 2 c 2 a3 b3 c3 thc P a 2b 3c b 2c 3a c 2a 3b Gii: Ta c: Trang 16
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1 . Tm gi tr nh nht ca biu
Vn dng bt ng thc tm GTLN - GTNN v gii phng trnh
P
a2
a4 2ab 3ac
b2
b4 2bc 3ba
c2
c4 2ca 3cb
(1)
p dng bt ng thc Bunhiacopski cho hai d y s sau:
a2 a2 2ab 3ac2
,2 2
b2 b2 2bc 3ba
,
c2 c2 2ca 3cb
ta c:
a
b
2
c
a2 . a2
a4 2ab 3ac
b2
b4 2bc 3ba
.c oc21
2ab 3ac b 2
2bc 3ba c 2
M a 2 b 2
c2P
1 , t (2) suy ra1 1 5 ab bc ca
t
h
a2 b2
c2
2
P a 2 b 2 c 2 5 ab bc ca
s
mc4 . 2ca 3cb 2ca 3cb(2) (3)
a2
2 ab 3ac , b 2
2bc 3ba , c 2
2ca 3cb v
Mt khc theo bt ng thc C si ta c:
a2 b2 b c2 2
2ab 2bc 2ca
a
2
ie t1
c
2
m3 3
a
ab bc ca
a2 b2 c2
T (3) ta c: P
1 5 ab bc caa b c
1 1 5.1
1 6
Du = xy ra Vy MinP =
1 6
w
Bi 5: Cho hai s dng a, b tha 0 a2 b2 1 a b thc M 1 a 1 b a b Gii:
w
w .v
a 1,0 b 1 . Tm gi tr nh nht ca biu
Ta c:
Trang 17
Vn dng bt ng thc tm GTLN - GTNN v gii phng trnh
M
p dng bt ng thc Bunhiacopski ta c:1 1 a 1 a 1 1 b 1 b 1 1 1 a 1 b 1 a b a b 1 a b 1 a2
1 b
1 a 1 b
a b1 1 a 1 1 b
ta b1
1 a b 1 a b
w .v
5 2 Bi ton tng qut: 2 a12 a2 Cho P 1 a1 1 a2 2n 1 Th minP nVy minM =
...
ie t
2 an 1 an
m
a
Du = xy ra
h1 3vi 0 ai
1
1
1
9 2
M
9 2 2
5 2
s1 i 1, n
a1 a2 ... an
Bi 6: Cho hm s thc f ( x)
x 2007
2009 x 2 . Tm gi tr ln nht v nh
nht ca f ( x ) trn min xc nh ca n. Gii:
Ta c min xc nh ca f ( x) : D Mt khc: f ( x)
w
2009; 2009
x 2007
2009 x 2
w
V f ( x) 0,x D
x Dx D
0; 2009x D
Do : max f ( x) Vi x D , ta c:
max f ( x) v min f ( x)2007. 2007 1. 2009 x 2
f ( x)
x
Theo bt ng thc Bunhiacopski th : Trang 18
.c oa b
f ( x)
max f ( x)x D
mf ( x) l hm l
a2 b2 1 1 a 1 b 2 1 a 1 b a b a2 1 a2 b2 1 b2 1 2 1 a 1 b a b 1 1 1 2 1 a 1 b a b
Vn dng bt ng thc tm GTLN - GTNN v gii phng trnh
2007. 2007 1. 2009
x2
2008 2007 2009 2008 4016 x 2
x2
x D
x D
Bi 7: Cho x, y, z thc T
0 tha mny2
xy
yz
t
x
y
y
z
z
x
Gii:
ie t
p dng bt ng thc Bunhiacopski ta c: 1 x y y z z x x y z y
m
a
x
2
z
2
hzx 1 . Tm gi tr nh nht ca biu zz z x
min f ( x)
2008 2008 ti x
2008
sxz z x
Vy max f ( x)
2008 2008 ti x
2008
.c ox y2T x
p dng bt ng thc Cosi ta c: x 2 4016 x 2 f ( x) 2008. 2008.2008 2 2007 1 2009 x 2 Du = xy ra x 2007 2 2 x 4016 x
2008
mz2
Suy ra: f ( x)
x 2008 4016 x 2
2008. x 2 4016
x2
x
y
z x2 x
2
x x y2
y
x
y
y y z y y z
y
z y
x z
w .vy y zy z x
z2
T
1 x 2
z x 1 2y z
x
Du = xy ra
w
1 3
Vy minT =
1 ti x 2
y
z
1 . 3
Bi 8: Cho ba s dng a, b, c tha mn: a b c 1 . Tm gi tr nh nht ca biu 1 1 1 1 thc P 2 2 2 a b c ab bc ca Gii:
w
Trang 19
Vn dng bt ng thc tm GTLN - GTNN v gii phng trnh
p dng bt ng thc Bunhiacopski ta c:
100
1 a2 b22
c2 1 ab
a
2
b
2
c
2
1 3 ab ab
1 3 bc bc
1 3 ca ca
2
a
1 b2
c
2 2
1 bc
1 ca
a2 b2
c 2 9ab 9bc 9ca
P a b c
7 ab bc ca
P 1 7 ab bc ca
a 2 b2 c2Do :
ab bc ca (suy ra t bt ng thc Cosi)2
P 30Du = xy ra Vy minP = 30 ti a
a
b
c
Bi ton tng qut: Cho n s dng a1 , a2 ,..., an n
ie t
b
c
1 3
w .v
t P =
1
a1 a2 ... a n
1 a1a2
m
1 3
a
100 P 1
7 a b c 3
10 P 3
2 v a1 a2 ... an 1 .1 a 2 a3 ... 1 an 1an 1 ana1
Th min P
n n
3
n
2
2
ta2 ... an
h
M ta li c: 1 2 a b c ab bc ca 3 Tht vy, t trn ta c: 2 a b c 3 ab bc ca
s1 n2
2
khi a1
2.3. S dng bt ng thc vect Lu : s dng bt ng thc vect th biu thc gi thit hoc biu thc cn tm gi tr ln nht, nh nht c dn g tng bnh phng ca cc s hng hoc cn bc hai ca tng bnh phng hoc l tng ca cc tch ca cc tha s . Bi 1: Cho hai s thc x, y tha mn 2 x 3 y 1 . Tm gi tr nh nht ca tng S 3x 2 2 y 2 Gii: Ta c S 3 x 2
w
w
2 y2
3x
2
2y
Trang 20
.c o
m
Vn dng bt ng thc tm GTLN - GTNN v gii phng trnh
Trong mt phng ta Oxy chn 2 3 4 9 35 u , u 3 2 6 3 2
v u.v
3 x, 2 y
v
3x 2
2 y2 35 . 3x 2 6 2 y2 3x 2 2 y2
2 x 3 y 1 u .v2 3x
Du = xy ra
4 y 9x
Kt hp vi iu kin ban u ta c: x Vy minS =
4 ,y 35
h
sx2 y2
6 ti x 35y
4 ,y 35
9 35
.c o9 35z2 a 1 b b 1 a2 a2 b2 2
3 2y
Bi 2: Cho x
z 1 . Tm gi tr nh nht ca biu thc P
Gii:
Trong mt phng ta Oxy chn: u x, y , z u x2 y2 z 2
m
vTa c: u.v
z , x, yu .v2 x2 3 x2 3 x2 x2
vxz xy
x2
a
y2
y2 y2 y2 y2
ie t
yz
x2z2 z2
t
z2y2 z2
z2
2 xz 2 xy 2 yz y2 y 2 xz 2 xy 2 yz 1 x
z2 z2
x2
w
1 3 x y z 1 Du = xy ra x y z z x y 3 1 1 Vy minP = khi x y z 3 3 2 2 Bi 3: Cho a b 1. Tm gi tr nh nht ca biu thc A z2
w
Trong mt phng ta Oxy chn: u a, b u a 2 b2 1
v
1 b, 1 a
w .v
Gii:
v
a b 2
Theo bt ng thc Bunhiacopski ta c: 1.a 1.b Trang 21
m6 35
Vn dng bt ng thc tm GTLN - GTNN v gii phng trnh
Do : v
2
2 u .v x y 2 2 2
A u.v
a 1 b b 1 a
Gii: Trong mt phng ta Oxy ta chn: 1 1 u x, u x2 x x2
v
y,
1 y
v
y2
1 y2
wu
z,
ww x
ie t
1 x
z2
1 z2
v
y
z,
1 x
p dng bt ng thc u
w .v
m
1 y
a
1 z
t
vz2
w1 z2
ux
h
Tm gi tr nh nht ca biu thc sau P
x2
1 x2
s
Bi 4: Cho ba s dng x, y, z v x
y
z 1.
.c oy2 1 y2 z2 1 z2
a b 1 b 1 a Kt hp vi iu kin ban u a 2 b 2 2 Suy ra: a b 2 2 Vy A max 2 2 khi a b 2Du = xy ra
1
vy
w ta c:z2
m1 x 1 y 1 z2
x
2
1 x2
y
2
1 y2
(1)
Nhn thy: x
y
z
2
w
1 x
1 y
1 z
2
81 x
y 1 x
z 1 y 1 y
2
80 x 1 z 1 z2
y
z
2
(2)
w
p dng bt ng thc Cosi ta c:
81 x
y
z
2
1 x
1 y
1 z
2
2 .9 x
y
z
1 x1 xyz
2.9.33 xyz .33T (2) v (3) ta c: Trang 22
2.81
(3)
Vn dng bt ng thc tm GTLN - GTNN v gii phng trnh
x
y
z
2
1 x
1 y
1 z
2
2.81 80
82
V do (1) nn:
P
x2
1 x2
y2
1 y2y
z2z
Du = xy ra khi v ch khi x Vy Pmin
82 khi x
y
z
1 . 3
Gii: Trong mt phng ta Oxy ta chn:
u v w
4a, ax 4b, by 4c, cz
u v w
16a 2 16b 2 16c 2
ax by
2
u v wTa c: u
4 a b c , ax by cz
ie t
m
2
cz
2
a
t8,6 u v w 10
v
w
u v w
w
c. Khng c vect no bng vect 0 a kb
w .v
P 16a 2 ax 16b 2 by 16c 2 Gi tr nh nht ca P: P min = 10 Du = xy ra khi v ch khi: a. C hai trong ba vect bng vect 0 b. C mt trong ba vect bng vect 0 Gi s u 0 th w k v k 0
2
2
h
P
16a 2
ax
2
16b 2
by
2
16c 2
scz2
Bi 5: Cho a b c
2 v ax by cz
6 . Tm gi tr nh nht ca biu thc
a b b c
ax by by cz
k
0
m 0
ax kby by mcz k, m 0 a b c 2 ax by cz 0
w
Trang 23
.c ocz2
1 z2 1 3
82
10
x y z 3 a b c 2 a , b, c 0
m
Vn dng bt ng thc tm GTLN - GTNN v gii phng trnh
Bi 6: Cho cc s dng x, y, z tha xy thc F x 4 y 4 z 4 Gii: Trong khng gian Oxyz chn: u x2 , y2 , z 2 u x4 y4
yz
zx
4 . Tm gi tr b nht ca biu
z4
v
1,1,12
v
3
Ta c: u.v M: u.v
x22
y22
z23 x4 y4 z4 x2 y2
u .v
y2 z2
z2
2 yz 2 xy yz
t
x 2 2 zx 2 x2 y2 z 2
hx2 y2 z 2 xy 16 3
s
Mt khc ta c: x 2 y 2 2 xy
.c oz22
mzx yz zx = 4
T ta c: 3 x 4
y4
z4
4 2 16
a2 1344a 82
x4
y4
z4
Vy: minF = 16 khi x
y
ie tu v w a2 5,5
Bi 7: Tm gi tr nh nht ca biu thc A a 2 4a 8 a 2 2ab b 2 Gii:
mz
b 2 6b 10
Trong mt phng ta Oxy chn:
u v
w .va 2,2 a b, 2
a b
4
w
b 3,1
b 2 6b 10 u v w 5 2
w
Ta c: u v w
w
u v w
u
v
w5 2
4a 8 a 2 2ab b 2 4 b 2 6b 10 a 2 2 b 3 Du = xy ra a 0, b 2 a 2 1 a bTrang 24
a2
Vn dng bt ng thc tm GTLN - GTNN v gii phng trnh
Vy A min
5 2 ti a
0, b
2
Bi 8: Cho a R . Tm gi tr nh nht ca biu thc M a 2 4a 13 a 2 2a 5 Gii:
Ta c: M a 2 9 a 1 Trong mt phng t Oxy chn:
2
2
4
u v
a 2,3 a 1,2 u v 3,5 v
u
a 2 a 12
2
9
4
u v
34
Bi 9: Cho ba s dng a, b, c tha: ab bc ca biu thc B
m
Vy: M min
34 khi a
1 5
a
Du = xy ra khi v ch khi a
1 5
tca
M: u
v
u v
a 2
2
9
ha 12
s4 34
abc . Tm gi tr nh nht ca
b2 ab
2a 2
ie t
c2
2b 2
a2
2c 2
bc
Gii: Ta c: B
1 2 1 2 1 2 2 2 2 a b b c c2 Trong mt phng ta Oxy chn: 1 2 1 2 u , u 2 a b a b2
w .v
2 a2
w
v
1 2 , b c 1 2 , c a
v w
1 b2 1 c2
2 c2 2 a2
w
w
1 1 1 1 , 2 c a b c 1 1 1 Mt khc: ab bc ca abc 1 a b cV u v
w
1 a
1 b
Trang 25
.c o
m
Vn dng bt ng thc tm GTLN - GTNN v gii phng trnh Do : u M: u
v w v w
1, 2 u v w
u v w
3
2.4. BI TP NGH
h1 x
Bi 1: Tm GTNN ca biu thc sau: a3 b3 c3 M 1 b 1 c 1 a 1 c 1 a 1 b Vi b 0, b 0, c 0 v abc 1 Bi 2: Tm GTLN ca hm s f ( x, y , z ) trn min D
s2 1 y 2 1 z
2
x, y , z : x
0, y
0, z
t2 x2
0 v x
w .v
M 5 x 2 2 xy 5y2 Bi 6: Tm gi tr nh nht ca hm s 2 2 x3 x12 x2 f ( x1 , x2 , x3 , x4 , x5 ) x2 x3 x4 x3 x4 x5 x4 x5 2 2 x5 x4 x5 x3 x2 x1 x2 x3Trn min D Bi 7: Cho x, y
ie t
Bai 3: Tm gi tr ln nht ca biu thc A ab bc 2ca vi a, b, c l cc s thc tha a 2 b 2 c 2 1 Bi 4: Tm gi tr ln nht ca biu thc P abc Trong a, b, c l cc s thc tha a 2 2b 2 2a 2c 2 b 2c 2 3a 2b 2c 2 9 Bi 5: Cho x 2 y 2 1 . Tm gi tr ln nht ca biu thc sau:
m
a
x1 , x2 , x3 , x4 , x5 : x12x2 9y2 4x2
2 x3
R . Tm gi tr nh nht ca biu thc sau:
w
Bi 8: Cho bit x 2 y 2 z 2 27 . Hy tm gi tr nh nht v ln nht ca hm s f ( x, y, z ) x y z xy yz zx Bi 9: Tm gi tr nh nht ca biu thc P a100 10a10 10 Bi 10: Cho x, y , z tha mn h sau:
w
A
9x
2
y2
y2
x2
x2
xy
y2
y 2 yz z 2 16 Tm gi tr ln nht ca biu thc sau: P xy yz
Trang 26
.c oy z 1
1 2 1 2 1 2 3 2 2 2 2 a b b c c2 a2 Du = xy ra khi v ch khi a b c 3 Vy B min 3 khi a b c 3 B
2 x4
2 x5 1
4y2
3zx
mx1
Vn dng bt ng thc tm GTLN - GTNN v gii phng trnh
Bi 11: Cho
a , b, c 0 . a b c 12 2 2
P 1 x 1 y 1 x 1 y Bi 14: Cho ba1 s thc a, b, c bt k. Tm gi tr nh nht ca biu thc2 2 2
p dng bt ng thc vect: u
v
u v
w
w
w .v
ie t
Bi 15: Cho x > 0, y > 0, z > 0 v x + y + z = 1. x y z Tm gi tr ln nht ca biu thc M = x 1 y 1 z 1 Hng dn: p dng bt ng thc Bunhiacopski cho hai d y: 1 1 1 v 1 x , 1 y , 1 z , , 1 x 1 y 1 z
m
Trang 27
a
t
h
P= b 1 c a b 1 c a Hng dn: Trong mt phng Oxy chn u b 1, c a , v b 1, a c
s
.c o2
F Bi 13: Cho a, b
a b b c b c 0,1 . Tm gi tr ln nht ca biu thc:
m
1 1 1 Tm gi tr nh nht ca biu thc P a b c a b c Bi 12: Cho a b c 1 v a, b, c 0 . Tm gi tr ln nht ca biu thc
Vn dng bt ng thc tm GTLN - GTNN v gii phng trnh
Phn 3: GII PHNG TRNH BNG PHNG PHP S DNG BT NG THC
3.1. Vn dng bt ng thc Csi
w .v
p dng bt ng thc Csi cho hai s khng m, ta c: 1 1 1 (1) 2 x8 x4 2x8 2 2x8 . 8 8 8 V
x4
1 4
ie t
Lu : p dng c bt ng thc Csi gii th: mt trong hai v ca phng trnh sau khi p dng bt ng thc Csi phi ln h n hoc bng (nh hn hoc bng) v cn li, hoc sau khi p dng bt ng thc th c mt ng thc c lng c nh hn (ln hn) hoc bng v cn li p dng c iu kin xy ra ca bt ng thc Csi . 3 Bi 1: Gii phng trnh: x 2 2 x 8 8 Gii:
2 x4.
1 4
m
x4
a
1 4
tx2
hx2x2 2
Ni v phng trnh th c rt nhiu loi phng trnh nh phng rnh bc hai, bc ba,phng trnh v t, phng trnh m, phng trnh logarit.Mi phng trnh c th c nhiu phng php gii khc nhau mu mc hay khng mu mc. Trong s cc phng php gii ca cc phng trnh th phng php s dng bt ng thc c th coi l phng php c o v sng to i hi ngi gii ton phi linh hot. S dng phng php ny ta c th s dng nhiu bt ng thc khc nhau, c th vn dng ring l hoc kt hp nhiu bt ng thc. Sau y l mt s bi ton gii phng trnh bng phng php vn dng bt ng thc m bt ng thc c s dng ch yu l bt ng thc Csi, Bunhiacopski v bt ng thc vect.
s2 x82 2
T (1) v (2) ta c: 2 x 8
x4
3 82x8 x4
x41 8 1 4
Ta c du = xy ra, do
w
Vy nghim ca phng trnh l x
w
Bi 2: Gii phng trnh: Gii:
x 3
5
x
x2
8 x 18
Trang 28
.c o3 8 x2
m(2)
Vn dng bt ng thc tm GTLN - GTNN v gii phng trnh
iu kin: 3 x 5 p dng bt ng thc Csi cho hai s khng m, ta c: x 3 5 x x 3 .1 5 x .1
x 42
2
22
x 4 0 x 4 Vy phng trnh cho c nghim l xBi 3: Gii cc phng trnh sau: a. x 2 4 x 5 2 2 x 3 x 1 3 b. 1 x 1 3 Gii:
2
4
a. x 2
4x 5
2 2x 3x
ie t
3 2 p dng bt ng thc Csi cho hai s khng m: 2 x 3 v 1, ta c: 2x 3 1 2 2x 3 x2 4x 5
2x 42
x2
4x 5
m
iu kin: 2 x 3 0
Du ng thc xy ra trong (1) khi v ch khi:
w
w
w .v
x 1 0 x 1 Th li x 1 l nghim ca phng trnh cho. Vy phng trnh cho c nghim duy nht l x x 1 3 (k: x b. 1) 1 x 1 3 p dng bt ng thc Csi , ta c: x 1 x 1 3 3 (1) 2 . 2 x 1 3 x 1 3
a
x2
t
2x 1 0
h1
sx x
.c o04
Do :
x 3
5
x
x
2
8 x 18
x 4
m22 22 4
x 3 1 5 x 1 2 2 2 Mt khc: x 2 8 x 18 x 2 8 x 16 2
x 1 x2 2x 1 9 x2 2x 8 x 1 3 Th li x 2 v x 4 l nghim ca phng trnh 3Vy phng trnh cho c nghim l x2 v x
Trang 29
Vn dng bt ng thc tm GTLN - GTNN v gii phng trnh
Bi 4: Gii phng trnh sau: 8 x 2
1 x
5 2
1 1 4 x
1 1 4 x
Gii:
w .v
0 1 x 0 x iu kin: x 1 1 1 0 x x 0 p dng bt ng thc Csi cho hai s khng m, ta c: 1 1 1 1 x 1 x .1 x 1. x x x x
x
ie t
Du = xy ra, ta c:
w
1 1 1 1 x 1 x 1 2 x 2 x 1 x 1 x x2 1 x 1 x 1 x 2 x x 1 0
m
a
Bi 5: Gii phng trnh sau: x
x
1 x
t1 1 x
h1 2
1 1 1 1 1 23 1 5 . . . . 5.5 8 x 2 . 2 4 4 x x x x 2 x 2 Du ng thc xy ra khi v ch khi: x 0 x 0 x 0 x 1 1 1 1 32 2.x 4 x5 8x 2 5 x 4 4 x Th li: x 4 tha mn Vy nghim ca phng trnh l x 4 55 8 x 2 .
s5
w
x
1 2
5
Kt hp vi iu kin ban u ta c: x Trang 30
.c o1 4
mx
Gii: iu kin: x 0 p dng bt ng thc Csi ta c: 1 1 1 1 1 8x 2 8x 2 x 4 x 4 x
Vn dng bt ng thc tm GTLN - GTNN v gii phng trnh
Vy nghim ca phng trnh l x Bi 6: Gii phng trnh sau: Gii: iu kin:
1 2
5
x2
4x 9
x2
4x 9
6
x2 2
4x 9 x2 9
x2 4xx4 2
4x 92
2 x4
x2 2x2
.c o4 x 9. x 2 2 811 3 3 2 3
x 2 x2 4x 9 0 x 2 5 0 p dng bt ng thc Csi cho hai s d ng, ta c:
x2
4x 9
0
x 2
2
5 0
R
9 x2iu kin: 3 x
0 0 0
m
Gii:
3 x
p dng bt ng thc Csi cho hai s khng m, ta c:
ie t
3
x
a
Bi 7: Gii phng trnh sau:
9 x2
3 x
t
h
Du = xy ra, do :
x 0 2x 0 Vy nghim ca phng trnh cho l: x
0
s0
2
3 x
3 2 3
3
3 x 3 x 2Do vy:
w .v
9 x2
3 x
3 x
3 x 3 x 1 3 x 3 . 2 3
3 x .3
m4x 9 2.3 6 811 3 3 x .3
1 3 x 3 . 2 3
3 x3 x 3 2 3x 0
3 x 3 3x 0
w
9 x
2
3 x
3 x 3 x
Vy phng trnh cho c nghim l:
w
Bi 8: Gii phng trnh sau: Gii:0
3
25 x 2 x 2
9
4x
3 x
iu kin: x
Trang 31
Vn dng bt ng thc tm GTLN - GTNN v gii phng trnh
Ta c:
3
25 x 2 x 23
9
4x9 9
25 x 4 2 x 2
3 x 4x25x 2
3 5x 2 2x2 9
3.3 25 x 4 2 x 2
p dng bt ng thc Csi cho ba s d ng: 5 x 2 ; 5 x 2 ; 2 x 2
m0
9 c:
Bi 9: Gii phng trnh 2 7 x 3 11x 2 Gii: Ta c: 2 7 x 3 11x 2
25 x 12
25 x 122
x2x
2 7x 4 x
x 3
a2
iu kin: 7 x 4 x 27x 4
x 3
m
0
t
6x 16x 1
h2
sx2x 1 22
0 (v x
x 3
4 7 p dng bt ng thc Csi cho hai s khng m: 7 x 4; x 2 x
ie t
.c o6x 12
(*) 5 x 2 5 x 2 2 x 2 9 3.3 25 x 4 2 x 2 9 Du = ng thc (*) xy ra khi v ch khi: 5x 2 2x 2 9 3x 2 9 x2 3 x 3 Th li: x 3 l nghim ca phng trnh cho 3 Vy phng trnh cho c nghim l x
11 4
x 3 c:
w .v
7x 4
x
2
x 3
2 7x 4 x
x 3
x2
6 x 1 2 7 x 3 11x 2
25 x 12
3.2. Vn dng bt ng thc BunhiacopskiLu : p dng c bt ng thc Bunhiacopski t h phng trnh phi c dng tch ca hai biu thc hoc tng ca cc biu thc m chng l tch ca hai tha s. V sau khi p dng bt ng thc Bunhiacopski th phi c phn a v
w
Du = xy ra khi v ch khi: 7 x 4 x 2 x 3 x 2 8x 7 0 x 1 (tha iu kin) x 7 Th li: x 1; x 7 l nghim Vy phng trnh cho c nghim l x 1; x 7
w
Trang 32
Vn dng bt ng thc tm GTLN - GTNN v gii phng trnh
biu thc gi thit ban u v a c v hng s. Sau vn dng iu kin bng nhau ca bt ng thc Bunhiacopski a ra nghim ca phng trnh. Bi 1: Gii phng trnh: Gii: iu kin: x 0 p dng bt ng thc Bunhiacopski cho hai cp s: x 1 c: 2 2 ; x 1 v ; x 1 x 1
ie t
(1) x 9 x 1 x 1 Du = trong (1) xy ra khi v ch khi: 1 2 2 2 2 1 x 1 2 2 1 x 1 . x 1 x x 1 x 1 x x 1 x x 1 8 1 1 (tha iu kin) 8x x 1 7 x 1 x x 1 x 7 1 Vy phng trnh c nghim l x 7
8
x 1 .
1
x
m
a
w
p dng bt ng thc Bunhiacopski cho cc cp s sau: 2; 3 v x 2 3 x 6; x 2 2 x 7 ta c:
w .v
Bi 2: Gii phng trnh: 13 x 2 3 x 6
2
t
h
2 2 x 1
x
1 2 2. x 1
s2
2
x 1.
.c o2
x
x 1
mx2 2x 7 5 x 2 12 x 332
2 2 x 1
x
x 9
Gii:
2 2 32 x 2 3 x 6
2
x2
2x 7
2 2 2
w
2 x 2 3x 6
3 x2
2x 7
5 x 2 12 x 33
Du = xy ra khi v ch khi: 3 x 2 3x 6 2 x 2 2 x 7
Trang 33
Vn dng bt ng thc tm GTLN - GTNN v gii phng trnh
x2 5x 4 0 x 1 x 4 Vy phng trnh cho c nghim l x 1; xBi 3: Gii phng trnh sau trn tp s N: 2 x 2 4 y 2 28 17 x 4 y 4 14 y 2 Gii: p dng bt ng thc Bunhiacopski ta c: 2 2 x 2 4 y 2 28 1. x 2 4 y 2 72 2
4
Bi 4: Gii phng trnh sau: Gii:
x2
m
1 42 x 2 y2 7 17 x 4 y 4 14 y 2 49 Du ng thc xy ra khi v ch khi : 4x2 y 2 7 2x y 2x y 7 V x, y N nn 2 x y 0 2x y 7 x 2 Ta c: 2x y 1 y 3 Vy phng trnh cho c nghim duy nht l x, y 2;3
a
t2x 1
h2
2x
s3x 23x 2
x2
2x
ie t
0
3x 2 4 x 1 0 p dng bt ng thc Bunhiacopski ta c:
w .v
iu kin: 2 x 1 0
x
1 2
x . x 2 1. 2 x 1
x2
12
x 2
.c o4x 1
49
2x 1
m2
w
Du ng thc trong (1) xy ra khi v ch khi: x . x 2 1. 2 x 1 2x2 x x 2
w
x 1 x 2 2x 1
x 1 3x 1
4x 1
(1)
x2
x 1 0
x x
1 2 1 2
5 5
Trang 34
Vn dng bt ng thc tm GTLN - GTNN v gii phng trnh
Kt hp iu kin ban u ta c nghim l x
1 25
5
Vy nghim ca phng trnh cho l x Bi 5: Gii phng trnh Gii: iu kin: 5 x 2 4 x 0 1 x 5 p dng bt ng thc Bunhiacopski ta c: 1 2 x 2 1. 5 x 2 4 x
1 2
5 x2
4x
2x 3 5
4
Bi 6: Gii phng trnh sau: 13 x 1 9 x 1 16 x Gii:
iu kin: x 1 p dng bt ng thc Bunhiacopski ta c: 13 x 1 9 x 1 13. 13 x 13 27 . 3 x 3
13 27 13 x 13 3 x 3 40 16 x 10 2 10 16 x 10 10 16 x 10 16 x (Bt ng thc Csi) ng thc xy ra khi v ch khi: 27 . 13 x 13 13. 3 x 3 5 (tha iu kin) x 4 10 16 x 10
w
w .v
ie t
m
a
5 x2 4x 2x 3 5 4 Du ng thc xy ra trong (2) 2 5 x2 x 2 0 6 5 x 2 2 5 5 x 20 x 16 0
t
h5 4
1
4 1. x 2
4x 4
5 x2
s4x4x1 x
Bi 7: Gii phng trnh sau: 4 x Gii:
w
Vy phng trnh cho c nghim l x4
1 x
x
.c o1 3 5 (2)x 224
Trang 35
m8
Vn dng bt ng thc tm GTLN - GTNN v gii phng trnh
x 0 0 x 1 1 x 0 p dng bt ng thc Bunhiacopski v Bunhiacopski m rng ta c: 1. x 1. 1 x 1 1 x 1 x 2iu kin:
1.4 x 1.4 1 x4
4
1 1 1 1 1 1 x 1 x
4
Vy nghim ca phng trnh cho l x Bi 8: Gii phng trnh sau: Gii: iu kin:
s
1 2
h
3x 2 1x 1 x
x2
x
x x2 1
.c o1 2 2 7x2 x 4(*)2 5x 2 x
x 41 x x 1 x 2 48 Du = trong ng thc xy ra khi v ch khi: x 1 x 1 (tha iu kin) x 4 4 2 x 1 x
3x 2 1 0 x2 x 0
3
x 2 1 1 3x 2 1 x 23x 2 1 x2 x Du = xy ra khi v ch
m
3 p dng bt ng thc Bunhiacopski m rng ta c: 1. 3 x 2 1 1. x 2 x x x 2 1
ie t
x x2 1
a
x
tx2 1x2
x x x
1 1 2 1 x 1(1)
3x 2 1 x2 1 x
x2
x
w .v
3x
2
1
x 1
Do (*) nn 5 x 2 x 0 p dng bt thc Csi ta c: 1 1 7x2 x 4 5x 2 x 2 x 2 2 2 2 2 2 1 .2 . 5 x 2 x .2 x 2 2 2 2 1 7x2 x 4 5 x 2 x .2 x 2 2 2 2 Du = xy ra khi v ch khi 5x 2 x 2 x 2 2 3x 2 x 4 0
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Trang 36
m8
Vn dng bt ng thc tm GTLN - GTNN v gii phng trnh
Bi 1: Gii phng trnh sau: Gii: iu kin: xR
x2
2x 5
x2
s2
Lu : p dng c bt ng thc vect vo vic gii phng trnh i hi phng trnh c cha cn bc hai ca hai tng b nh phng ta phn tch thnh ln vect, hoc cha tng ca hai tch cho thy c s phn tch ca tch v hng ca hai vect. T ta p dng cc bt ng thc vect bit c lng v vn dng iu kin xy ra ca du = t m nghim ca phng trnh.
6 x 10
w u
u v v
2,1 x 12
w 4
5
ie t
x 3
m
Ta vit li phng trnh: x 1 4 x 3 1 5 Trong mt phng ta Oxy chn cc vect c ta sau: u x 1,2 ; v x 3,1
a
2
t
h
.c ou
3.3. Vn dng bt ng thc vect
2
1k v vi k >0
Do (*) nn: u v Nn:
u
v , du = xy ra
Vy nghim ca phng trnh l x Bi 2: Gii phng trnh sau:
w .v
x 1 (iu kin: x 0 ) 2 x 3 x 1 2x 6 x 5 (tha iu kin)5
x2
2x
10
x2
6x
m5(*)
1 4 x 3 T (1) v (2) ta c nghim ca phng trnh l: x Vy nghim ca phng trnh cho l x 1
x
(2)
1
13
41
w
Gii:R2 2
w
iu kin: x
Ta vit li phng trnh: x 1 9 3 x 4 41 Trong mt phng ta Oxy chn cc vect c ta sau:
(*)
u v
x 1,3 3 x, 2
u v
x 1
2
92
3 x
4
Trang 37
Vn dng bt ng thc tm GTLN - GTNN v gii phng trnh
u v
4,5
u v
41 u v ux
Kt hp vi (*) nn: u v Do :
kv
Vy nghim ca phng trnh l x Bi 3: Gii phng trnh: 3 x x 1 Gii: 5 iu kin: 1 x 2
5 2x
40 34 x 10 x 2
.c ox3x32003
7 5
m(1) (2)
x 1 3 x
3 2
2x 2
9 3x
7 5
(1)
3 x
x 1
5 2x
3 x
Trong mt phng ta Oxy chn cc vect c ta sau:
u vu.v
3 x,1
u
3 x v
1
3 x
x 12
5 2x
V (2) nn: Do :
u.v
ie tu .vu
u .v
3 x
1. 4 x
mkv
x 1, 5 2 x
4 x
40 34 x 10 x 2(k>0)
w .v
5 3 x 1 (iu kin: x ; x 1) 2 x 1 5 2x 2 x 3 17 x 2 49 x 46 0 x 2 (tha iu kin) Vy nghim ca phng trnh l x = 2Bi 4: Gii phng trnh sau:
x2
ax 2 10 x 267
2
8 x 816
w
Gii: iu kin: x R Trong mt phng ta Oxy chn
w
u v
4 x,20 2
u v u v
x2
8 x 816
5 x,11 2 9,31 2
x 2 10 x 267 81 2.312u v
u v
t
Theo bi ta c: u v
Trang 38
h2
s4 4 x2003
Vn dng bt ng thc tm GTLN - GTNN v gii phng trnh
Du = xy ra
4 x 5 x
20 2 11 2
44 11x 100 20 x
31x
56
x
Vy nghim ca phng trnh l: x Bi 5: Gii phng trnh sau: Gii: iu kin: 2 x 4 Trong mt phng ta Oxy chn:
x 2
4 x
x2
v
x 2, 4 x
v
2
u .vx 2
2, u.v4 x
x 2
4 x
a
t1 x 22
h1 4 x2
u
1,1
u
2
2 du = xy ra
Nhn thy: x 2
6 x 11
x 2 Vy nghim ca phng trnh l: xBi 6: Gii phng trnh sau:
ie t
Do du = xy ra
x 3 x 3
m
2
s4 x 3 x 31 2x 1 2x
Gii: 1 1 Diu kin: x 2 2 Trong mt phng ta Oxy chn:
v
w
u
1,1
1 2x , 1 2x1 2x 2
w .v
1 2x
1 2x
1 2x 1 2x
u
2
v
2
w
u.v u .v
1 2x
M: u.v
u .v
1 2x
1 2x
2
Nhn thy:
1 2x 1 2x
1 2x 1 2x
2
1 2x 1 2x . 1 2x 1 2x
Trang 39
.c o6 x 11
2 (BT Csi)
m
56 31 56 31
Vn dng bt ng thc tm GTLN - GTNN v gii phng trnh
1 2xDu = xy ra
1 2x 1 2x 1 2x x 0
1 2x 1 2x 1 2x 1 2x Vy nghim ca phng trnh l: x = 0Bi 7: Gii phng trnh sau: Gii: iu kin: x 4 Ta vit li phng trnh di dng sau:x 2 x 4 3
x 4 x 4
u v 2
x 4 1,0 x 4 ,0 1,0v
u v 1 2
x 4 1 x 4
2
u vM: u
u vu v
Theo (*) du = xy ra
m
ie t
4 x 1 2 4 x4 x 1 2
a
t
2
h0R
1
w .v
x 4 1 2 1 2 x 4 1 1 2 1 1 2 12
x 4 x 4
w
Vy nghim ca phng trnh l: x
4
1 2 1
sx 42
x 4 1 + 2 x 4 = 1 (*) Trong mt phng ta Oxy chn:
2
2
.c o2
vi
w
Bi 8: Gii phng trnh sau: Gii:
x 1 x 3
2x 3
2
2x 1
iu kin: x 1 Trong mt phng ta Oxy chn: Trang 40
m1
R
Vn dng bt ng thc tm GTLN - GTNN v gii phng trnh
u v u.v u .v
x 1, x 3 1,1 v x 1 2 x 32
u
x 1
x 3
2
2x 3
2x 1
Du = xy ra
x 1 1 x 3
x 3 1
x x 32
3 x 3
x2
7 x 10
0
2)
x 3 4 x 1
ie t
Gii cc phng trnh sau: 3 1) 4 x 2x 8 3) x 4 4) x 2 5)
x 8 6 x 1
m
3.4. BI TP NGH
4
2 x4
4
2 x4
4
2x 44x 9
3 x3
4x
w .v
x2
x2
4x 9
x3 2
6) x 3 3 x 2 8 x 40 2x 3 7) x 1 8)
84 4 x 4 50 3 x 12
x22
2x 2 2x 5
x22
2x 2
2 2 29
w
9) x x 1 10) x
3x
2 xx
2
1
2 x 10
w
Trang 41
a
x 5 Vy nghim ca phng trnh l x = 5
6
t1
hx x
x
3
s2 5
x 1
x 1
.c ox2 6x 9
Ta c: u.v
u .v
m
Vn dng bt ng thc tm GTLN - GTNN v gii phng trnh
KT LUNCc dng ton lin quan n bt ng thc thng khng d nn cc dng ton ny thng ch s dng tuyn chn cc hc sinh gii. Ban u, n ch bit di dng chng minh cc bt ng thc tr n c s cc bt ng thc thng dng, nhng sau cc dng ton ra i trn c s cc bt ng thc thng dng bit nh: tm gi tr ln nht, gi tr nh nht, gii ph ng trnh, h phng trnh, bt phng trnh v h bt phng trnh.
Trong ti ny ti ch nghin cu hai dng ton l tm gi tr ln nht, gi
- p dng c cc bt ng thc gii ton i hi k nng nhn xt ca
- Mc d cc dng ton v bt ng thc rt kh, kh nht l a v ng dng bt
tun th cc nguyn tc bin i ng thc nhn xt nhy bn a v dng ca bt ng thc cn ng dng th bi ton s tr nn khng kh. Qua ti ti hc c rt nhiu kinh nghim trong gii ton bt ng thc v thy c mi lin h ca cc bt ng thc vi nhau.
w
w
w .v
ie t
m
ng thc cn vn dng nhng khi ta bit s dng thnh tho cc bt ng thc v
Trang 42
a
hot a v ng dng ca bt ng thc cn p dng.
t
ngi gii phi nhy bn, v k nng bin i tng ng cc biu thc phi linh
h
v bt ng thc vect. Qua qu trnh thc hin ti rt ra c cc iu sau:
s
tr nh nht v gii phng trnh da trn ba bt ng thc l: Csi, Bunhiacopski
.c o
m
Vn dng bt ng thc tm GTLN - GTNN v gii phng trnh
TI LIU THAM KHO1) Phan Huy Khi, Chuyn bi dng hc sinh gii Ton THCS: Gi tr ln nht v nh nht ca hm s, NXB Gio Dc, nm 2008.
2) T sch Ton hc & tui tr, Cc b i thi Olympic Ton THPT (1990
3) V Giang Giai, Chuyn Bt ng Thc, NXB HQG H Ni, nm 2002. 4) Nguyn Th Hng, Bt ng thc v bt phng trnh i s, NXB HQG T.P H Ch Minh, nm 2003.
5) H Vn Chng, Tuyn tp 700 bi ton bt ng thc luyn thi v o cc
6) Nguyn c Tun, Nguyn Anh Ho ng, Trn Vn Hnh, Nguyn o n V, Gii phng trnh bt phng trnh h phng trnh h bt phng
7) Trn nh Th, Dng hnh hc gii tch gii phng trnh bt phng trnh h phng trnh bt ng thc....,NXB HQG H Ni, nm 2008. 8) Trn Vn K, Chn lc 394 bi ton bt ng thc gi tr ln nht gi tr nh nht, NXB T.P H Ch Minh, nm 2002.
w
w
w .v
ie t
m
Trang 43
a
trnh bng bt ng thc, NXB HQG T.P H Ch Minh, nm 2006.
t
h
trng H C bi dng hc sinh gii PTTH, NXB Tr, nm 1993.
s
.c o
2000).
m
Vn dng bt ng thc tm GTLN - GTNN v gii phng trnh
PH LCCc bi ton tm gi tr ln nht, gi tr nh nht trong cc thi i hc gn y. 1. (Khi A nm 2006) Cho hai s thc x 0, y 0 thay i v tho mn iu kin: 1 1 x y xy x 2 y 2 xy . Tm gi tr ln nht ca biu thc A 3 x y3 2. (Khi B nm 2006) Cho x, y l cc s thc thay i. Tm gi tr nh nht ca biu thc
w
w
w .v
x 1 y 1 z 1 y z 2 yz 2 zx 2 xy 5. (Khi B nm 2008) Cho hai s thc x, y thay i tho mn h thc x 2 y 2 1 . Tm gi tr ln 2 x 2 6 xy nht v gi tr nh nht ca biu thc P 1 2 xy 2 y 2 6. (Khi D nm 2008) Cho x, y l hai s thc khng m thay i. T m gi tr ln nht v gi tr nh x y 1 xy nht ca biu thc P 2 2 1 x 1 yP
x
ie t
m
Trang 44
a
t
x 1 y2 x 1 y2 y 2 A 3. (Khi A nm 2007) Cho x, y, z l cc s thc dng thay i v tho mn iu kin: xyz 1 . Tm x2 y z y2 z x z2 x y gi tr nh nht ca biu thc P = y y 2z z z z 2x x x x 2 y y 4. (Khi B nm 2007) Cho x, y , z l cc s thc dng thay i. Tm gi tr nh nht ca biu thc
2
2
h
s
.c o
m
Vn dng bt ng thc tm GTLN - GTNN v gii phng trnh
MC LCPHN M U ................................................................................................................... 1 I. L DO CHN TI ............................................................................................. 2 II. MC CH NGHIN CU...................................................................................... 2 III. I TNG NGHIN CU ..................................................................................... 2 IV. PHM VI NGHIN CU .......................................................................................... 2 VI. PHNG PHP NGHIN C U ......................................................................... 2 PHN NI DUNG ................................................................................................................ 3 Phn 1: S LC V BT NG THC ..................................................................... 4 1.1. nh ngha bt ng thc ................................................................................... 4 1.2. Tnh cht c bn ca bt ng thc ....................................................................... 4 1.3. Mt s bt ng thc c bn................................................................................... 4 Phn 2: TM GI TR LN NHT V GI TR NH NHT ...................................... 6 CA HM S HOC BIU THC ................................................................................ 6 2.1 KIN THC CN NH ......................................................................................... 6 2.1.1. nh ngha........................................................................................................ 6 2.1.2. Tm gi tr nh nht, gi tr ln nht ca biu thc (h m s) bng phng php vn dng bt ng thc ..................................................................................... 6 2.2. BI TP ................................................................................................................. 7 2.2.1. S dng bt ng thc Csi ............................................................................. 7 2.2.2. S dng bt ng thc Bunhi acopski ............................................................ 15 2.3. S dng bt ng thc vect ................................................................................. 20 2.4. BI TP NGH.............................................................................................. 26 Phn 3: GII PHNG TRNH BNG PHNG PHP ........................................... 28 S DNG BT NG THC ...................................................................................... 28 3.1. Vn dng bt ng thc Csi ................................................................................ 28 3.2. Vn dng bt ng thc Bunhiacopski .................................................................32 3.3. Vn dng bt ng thc vect .............................................................................. 37 3.4. BI TP NGH.............................................................................................. 41 KT LUN...................................................................................................................... 42 TI LIU THAM KHO ............................................................................................... 43 PH LC ........................................................................................................................ 44 MC LC ....................................................................................................................... 45
w
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w .v
ie t
m
Trang 45
a
t
h
s
.c o
m